3.432 \(\int \frac{1}{(c+\frac{a}{x^2}+\frac{b}{x})^3 x} \, dx\)

Optimal. Leaf size=190 \[ \frac{b \left (30 a^2 c^2-10 a b^2 c+b^4\right ) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{c^3 \left (b^2-4 a c\right )^{5/2}}-\frac{b x \left (b^2-7 a c\right )}{c^2 \left (b^2-4 a c\right )^2}+\frac{x^4 (2 a+b x)}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}+\frac{x^2 \left (b x \left (b^2-10 a c\right )+a \left (b^2-16 a c\right )\right )}{2 c \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}+\frac{\log \left (a+b x+c x^2\right )}{2 c^3} \]

[Out]

-((b*(b^2 - 7*a*c)*x)/(c^2*(b^2 - 4*a*c)^2)) + (x^4*(2*a + b*x))/(2*(b^2 - 4*a*c)*(a + b*x + c*x^2)^2) + (x^2*
(a*(b^2 - 16*a*c) + b*(b^2 - 10*a*c)*x))/(2*c*(b^2 - 4*a*c)^2*(a + b*x + c*x^2)) + (b*(b^4 - 10*a*b^2*c + 30*a
^2*c^2)*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/(c^3*(b^2 - 4*a*c)^(5/2)) + Log[a + b*x + c*x^2]/(2*c^3)

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Rubi [A]  time = 0.27942, antiderivative size = 190, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.444, Rules used = {1354, 738, 818, 773, 634, 618, 206, 628} \[ \frac{b \left (30 a^2 c^2-10 a b^2 c+b^4\right ) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{c^3 \left (b^2-4 a c\right )^{5/2}}-\frac{b x \left (b^2-7 a c\right )}{c^2 \left (b^2-4 a c\right )^2}+\frac{x^4 (2 a+b x)}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}+\frac{x^2 \left (b x \left (b^2-10 a c\right )+a \left (b^2-16 a c\right )\right )}{2 c \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}+\frac{\log \left (a+b x+c x^2\right )}{2 c^3} \]

Antiderivative was successfully verified.

[In]

Int[1/((c + a/x^2 + b/x)^3*x),x]

[Out]

-((b*(b^2 - 7*a*c)*x)/(c^2*(b^2 - 4*a*c)^2)) + (x^4*(2*a + b*x))/(2*(b^2 - 4*a*c)*(a + b*x + c*x^2)^2) + (x^2*
(a*(b^2 - 16*a*c) + b*(b^2 - 10*a*c)*x))/(2*c*(b^2 - 4*a*c)^2*(a + b*x + c*x^2)) + (b*(b^4 - 10*a*b^2*c + 30*a
^2*c^2)*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/(c^3*(b^2 - 4*a*c)^(5/2)) + Log[a + b*x + c*x^2]/(2*c^3)

Rule 1354

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + 2*n*p)*(c + b/x^n +
a/x^(2*n))^p, x] /; FreeQ[{a, b, c, m, n}, x] && EqQ[n2, 2*n] && ILtQ[p, 0] && NegQ[n]

Rule 738

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(
d*b - 2*a*e + (2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] + Dist[1/((p + 1)*(b^2 -
4*a*c)), Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2*c*d^2*(2*p + 3) + e*(b*e - 2*d*
c)*(m + 2*p + 2)*x, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] &
& NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, b, c, d,
e, m, p, x]

Rule 818

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1)*(2*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (2*c^2*d*f + b^2*e*g
- c*(b*e*f + b*d*g + 2*a*e*g))*x))/(c*(p + 1)*(b^2 - 4*a*c)), x] - Dist[1/(c*(p + 1)*(b^2 - 4*a*c)), Int[(d +
e*x)^(m - 2)*(a + b*x + c*x^2)^(p + 1)*Simp[2*c^2*d^2*f*(2*p + 3) + b*e*g*(a*e*(m - 1) + b*d*(p + 2)) - c*(2*a
*e*(e*f*(m - 1) + d*g*m) + b*d*(d*g*(2*p + 3) - e*f*(m - 2*p - 4))) + e*(b^2*e*g*(m + p + 1) + 2*c^2*d*f*(m +
2*p + 2) - c*(2*a*e*g*m + b*(e*f + d*g)*(m + 2*p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && Ne
Q[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && ((EqQ[m, 2] && EqQ[p, -3] &&
RationalQ[a, b, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rule 773

Int[(((d_.) + (e_.)*(x_))*((f_) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*g*x)/
c, x] + Dist[1/c, Int[(c*d*f - a*e*g + (c*e*f + c*d*g - b*e*g)*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c,
d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{1}{\left (c+\frac{a}{x^2}+\frac{b}{x}\right )^3 x} \, dx &=\int \frac{x^5}{\left (a+b x+c x^2\right )^3} \, dx\\ &=\frac{x^4 (2 a+b x)}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}-\frac{\int \frac{x^3 (8 a+b x)}{\left (a+b x+c x^2\right )^2} \, dx}{2 \left (b^2-4 a c\right )}\\ &=\frac{x^4 (2 a+b x)}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}+\frac{x^2 \left (a \left (b^2-16 a c\right )+b \left (b^2-10 a c\right ) x\right )}{2 c \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}-\frac{\int \frac{x \left (2 a \left (b^2-16 a c\right )+2 b \left (b^2-7 a c\right ) x\right )}{a+b x+c x^2} \, dx}{2 c \left (b^2-4 a c\right )^2}\\ &=-\frac{b \left (b^2-7 a c\right ) x}{c^2 \left (b^2-4 a c\right )^2}+\frac{x^4 (2 a+b x)}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}+\frac{x^2 \left (a \left (b^2-16 a c\right )+b \left (b^2-10 a c\right ) x\right )}{2 c \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}-\frac{\int \frac{-2 a b \left (b^2-7 a c\right )+\left (2 a c \left (b^2-16 a c\right )-2 b^2 \left (b^2-7 a c\right )\right ) x}{a+b x+c x^2} \, dx}{2 c^2 \left (b^2-4 a c\right )^2}\\ &=-\frac{b \left (b^2-7 a c\right ) x}{c^2 \left (b^2-4 a c\right )^2}+\frac{x^4 (2 a+b x)}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}+\frac{x^2 \left (a \left (b^2-16 a c\right )+b \left (b^2-10 a c\right ) x\right )}{2 c \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}+\frac{\int \frac{b+2 c x}{a+b x+c x^2} \, dx}{2 c^3}-\frac{\left (b \left (b^4-10 a b^2 c+30 a^2 c^2\right )\right ) \int \frac{1}{a+b x+c x^2} \, dx}{2 c^3 \left (b^2-4 a c\right )^2}\\ &=-\frac{b \left (b^2-7 a c\right ) x}{c^2 \left (b^2-4 a c\right )^2}+\frac{x^4 (2 a+b x)}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}+\frac{x^2 \left (a \left (b^2-16 a c\right )+b \left (b^2-10 a c\right ) x\right )}{2 c \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}+\frac{\log \left (a+b x+c x^2\right )}{2 c^3}+\frac{\left (b \left (b^4-10 a b^2 c+30 a^2 c^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{b^2-4 a c-x^2} \, dx,x,b+2 c x\right )}{c^3 \left (b^2-4 a c\right )^2}\\ &=-\frac{b \left (b^2-7 a c\right ) x}{c^2 \left (b^2-4 a c\right )^2}+\frac{x^4 (2 a+b x)}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}+\frac{x^2 \left (a \left (b^2-16 a c\right )+b \left (b^2-10 a c\right ) x\right )}{2 c \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}+\frac{b \left (b^4-10 a b^2 c+30 a^2 c^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{c^3 \left (b^2-4 a c\right )^{5/2}}+\frac{\log \left (a+b x+c x^2\right )}{2 c^3}\\ \end{align*}

Mathematica [A]  time = 0.337448, size = 221, normalized size = 1.16 \[ \frac{\frac{-39 a^2 b^2 c^2+50 a^2 b c^3 x+32 a^3 c^3-30 a b^3 c^2 x+11 a b^4 c+4 b^5 c x-b^6}{\left (b^2-4 a c\right )^2 (a+x (b+c x))}+\frac{a^2 b c (5 c x-4 b)+2 a^3 c^2+a b^3 (b-5 c x)+b^5 x}{\left (b^2-4 a c\right ) (a+x (b+c x))^2}-\frac{2 b c \left (30 a^2 c^2-10 a b^2 c+b^4\right ) \tan ^{-1}\left (\frac{b+2 c x}{\sqrt{4 a c-b^2}}\right )}{\left (4 a c-b^2\right )^{5/2}}+c \log (a+x (b+c x))}{2 c^4} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((c + a/x^2 + b/x)^3*x),x]

[Out]

((-b^6 + 11*a*b^4*c - 39*a^2*b^2*c^2 + 32*a^3*c^3 + 4*b^5*c*x - 30*a*b^3*c^2*x + 50*a^2*b*c^3*x)/((b^2 - 4*a*c
)^2*(a + x*(b + c*x))) + (2*a^3*c^2 + b^5*x + a*b^3*(b - 5*c*x) + a^2*b*c*(-4*b + 5*c*x))/((b^2 - 4*a*c)*(a +
x*(b + c*x))^2) - (2*b*c*(b^4 - 10*a*b^2*c + 30*a^2*c^2)*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]])/(-b^2 + 4*a*c
)^(5/2) + c*Log[a + x*(b + c*x)])/(2*c^4)

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Maple [B]  time = 0.013, size = 530, normalized size = 2.8 \begin{align*}{\frac{1}{ \left ( c{x}^{2}+bx+a \right ) ^{2}} \left ({\frac{b \left ( 25\,{a}^{2}{c}^{2}-15\,a{b}^{2}c+2\,{b}^{4} \right ){x}^{3}}{{c}^{2} \left ( 16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4} \right ) }}+{\frac{ \left ( 32\,{a}^{3}{c}^{3}+11\,{a}^{2}{b}^{2}{c}^{2}-19\,a{b}^{4}c+3\,{b}^{6} \right ){x}^{2}}{2\,{c}^{3} \left ( 16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4} \right ) }}+{\frac{ab \left ( 31\,{a}^{2}{c}^{2}-22\,a{b}^{2}c+3\,{b}^{4} \right ) x}{{c}^{3} \left ( 16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4} \right ) }}+{\frac{3\,{a}^{2} \left ( 8\,{a}^{2}{c}^{2}-7\,a{b}^{2}c+{b}^{4} \right ) }{2\,{c}^{3} \left ( 16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4} \right ) }} \right ) }+8\,{\frac{\ln \left ( c{x}^{2}+bx+a \right ){a}^{2}}{c \left ( 16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4} \right ) }}-4\,{\frac{\ln \left ( c{x}^{2}+bx+a \right ) a{b}^{2}}{{c}^{2} \left ( 16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4} \right ) }}+{\frac{\ln \left ( c{x}^{2}+bx+a \right ){b}^{4}}{2\,{c}^{3} \left ( 16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4} \right ) }}-30\,{\frac{{a}^{2}b}{c \left ( 16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4} \right ) \sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+10\,{\frac{a{b}^{3}}{{c}^{2} \left ( 16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4} \right ) \sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }-{\frac{{b}^{5}}{{c}^{3} \left ( 16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4} \right ) }\arctan \left ({(2\,cx+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c+a/x^2+b/x)^3/x,x)

[Out]

(1/c^2*b*(25*a^2*c^2-15*a*b^2*c+2*b^4)/(16*a^2*c^2-8*a*b^2*c+b^4)*x^3+1/2*(32*a^3*c^3+11*a^2*b^2*c^2-19*a*b^4*
c+3*b^6)/c^3/(16*a^2*c^2-8*a*b^2*c+b^4)*x^2+a*b*(31*a^2*c^2-22*a*b^2*c+3*b^4)/(16*a^2*c^2-8*a*b^2*c+b^4)/c^3*x
+3/2*a^2*(8*a^2*c^2-7*a*b^2*c+b^4)/c^3/(16*a^2*c^2-8*a*b^2*c+b^4))/(c*x^2+b*x+a)^2+8/c/(16*a^2*c^2-8*a*b^2*c+b
^4)*ln(c*x^2+b*x+a)*a^2-4/c^2/(16*a^2*c^2-8*a*b^2*c+b^4)*ln(c*x^2+b*x+a)*a*b^2+1/2/c^3/(16*a^2*c^2-8*a*b^2*c+b
^4)*ln(c*x^2+b*x+a)*b^4-30/c/(16*a^2*c^2-8*a*b^2*c+b^4)/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*
a^2*b+10/c^2/(16*a^2*c^2-8*a*b^2*c+b^4)/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*a*b^3-1/c^3/(16*
a^2*c^2-8*a*b^2*c+b^4)/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*b^5

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c+a/x^2+b/x)^3/x,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.05508, size = 3429, normalized size = 18.05 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c+a/x^2+b/x)^3/x,x, algorithm="fricas")

[Out]

[1/2*(3*a^2*b^6 - 33*a^3*b^4*c + 108*a^4*b^2*c^2 - 96*a^5*c^3 + 2*(2*b^7*c - 23*a*b^5*c^2 + 85*a^2*b^3*c^3 - 1
00*a^3*b*c^4)*x^3 + (3*b^8 - 31*a*b^6*c + 87*a^2*b^4*c^2 - 12*a^3*b^2*c^3 - 128*a^4*c^4)*x^2 + (a^2*b^5 - 10*a
^3*b^3*c + 30*a^4*b*c^2 + (b^5*c^2 - 10*a*b^3*c^3 + 30*a^2*b*c^4)*x^4 + 2*(b^6*c - 10*a*b^4*c^2 + 30*a^2*b^2*c
^3)*x^3 + (b^7 - 8*a*b^5*c + 10*a^2*b^3*c^2 + 60*a^3*b*c^3)*x^2 + 2*(a*b^6 - 10*a^2*b^4*c + 30*a^3*b^2*c^2)*x)
*sqrt(b^2 - 4*a*c)*log((2*c^2*x^2 + 2*b*c*x + b^2 - 2*a*c + sqrt(b^2 - 4*a*c)*(2*c*x + b))/(c*x^2 + b*x + a))
+ 2*(3*a*b^7 - 34*a^2*b^5*c + 119*a^3*b^3*c^2 - 124*a^4*b*c^3)*x + (a^2*b^6 - 12*a^3*b^4*c + 48*a^4*b^2*c^2 -
64*a^5*c^3 + (b^6*c^2 - 12*a*b^4*c^3 + 48*a^2*b^2*c^4 - 64*a^3*c^5)*x^4 + 2*(b^7*c - 12*a*b^5*c^2 + 48*a^2*b^3
*c^3 - 64*a^3*b*c^4)*x^3 + (b^8 - 10*a*b^6*c + 24*a^2*b^4*c^2 + 32*a^3*b^2*c^3 - 128*a^4*c^4)*x^2 + 2*(a*b^7 -
12*a^2*b^5*c + 48*a^3*b^3*c^2 - 64*a^4*b*c^3)*x)*log(c*x^2 + b*x + a))/(a^2*b^6*c^3 - 12*a^3*b^4*c^4 + 48*a^4
*b^2*c^5 - 64*a^5*c^6 + (b^6*c^5 - 12*a*b^4*c^6 + 48*a^2*b^2*c^7 - 64*a^3*c^8)*x^4 + 2*(b^7*c^4 - 12*a*b^5*c^5
+ 48*a^2*b^3*c^6 - 64*a^3*b*c^7)*x^3 + (b^8*c^3 - 10*a*b^6*c^4 + 24*a^2*b^4*c^5 + 32*a^3*b^2*c^6 - 128*a^4*c^
7)*x^2 + 2*(a*b^7*c^3 - 12*a^2*b^5*c^4 + 48*a^3*b^3*c^5 - 64*a^4*b*c^6)*x), 1/2*(3*a^2*b^6 - 33*a^3*b^4*c + 10
8*a^4*b^2*c^2 - 96*a^5*c^3 + 2*(2*b^7*c - 23*a*b^5*c^2 + 85*a^2*b^3*c^3 - 100*a^3*b*c^4)*x^3 + (3*b^8 - 31*a*b
^6*c + 87*a^2*b^4*c^2 - 12*a^3*b^2*c^3 - 128*a^4*c^4)*x^2 + 2*(a^2*b^5 - 10*a^3*b^3*c + 30*a^4*b*c^2 + (b^5*c^
2 - 10*a*b^3*c^3 + 30*a^2*b*c^4)*x^4 + 2*(b^6*c - 10*a*b^4*c^2 + 30*a^2*b^2*c^3)*x^3 + (b^7 - 8*a*b^5*c + 10*a
^2*b^3*c^2 + 60*a^3*b*c^3)*x^2 + 2*(a*b^6 - 10*a^2*b^4*c + 30*a^3*b^2*c^2)*x)*sqrt(-b^2 + 4*a*c)*arctan(-sqrt(
-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)) + 2*(3*a*b^7 - 34*a^2*b^5*c + 119*a^3*b^3*c^2 - 124*a^4*b*c^3)*x + (a
^2*b^6 - 12*a^3*b^4*c + 48*a^4*b^2*c^2 - 64*a^5*c^3 + (b^6*c^2 - 12*a*b^4*c^3 + 48*a^2*b^2*c^4 - 64*a^3*c^5)*x
^4 + 2*(b^7*c - 12*a*b^5*c^2 + 48*a^2*b^3*c^3 - 64*a^3*b*c^4)*x^3 + (b^8 - 10*a*b^6*c + 24*a^2*b^4*c^2 + 32*a^
3*b^2*c^3 - 128*a^4*c^4)*x^2 + 2*(a*b^7 - 12*a^2*b^5*c + 48*a^3*b^3*c^2 - 64*a^4*b*c^3)*x)*log(c*x^2 + b*x + a
))/(a^2*b^6*c^3 - 12*a^3*b^4*c^4 + 48*a^4*b^2*c^5 - 64*a^5*c^6 + (b^6*c^5 - 12*a*b^4*c^6 + 48*a^2*b^2*c^7 - 64
*a^3*c^8)*x^4 + 2*(b^7*c^4 - 12*a*b^5*c^5 + 48*a^2*b^3*c^6 - 64*a^3*b*c^7)*x^3 + (b^8*c^3 - 10*a*b^6*c^4 + 24*
a^2*b^4*c^5 + 32*a^3*b^2*c^6 - 128*a^4*c^7)*x^2 + 2*(a*b^7*c^3 - 12*a^2*b^5*c^4 + 48*a^3*b^3*c^5 - 64*a^4*b*c^
6)*x)]

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Sympy [B]  time = 2.83448, size = 1510, normalized size = 7.95 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c+a/x**2+b/x)**3/x,x)

[Out]

(-b*sqrt(-(4*a*c - b**2)**5)*(30*a**2*c**2 - 10*a*b**2*c + b**4)/(2*c**3*(1024*a**5*c**5 - 1280*a**4*b**2*c**4
+ 640*a**3*b**4*c**3 - 160*a**2*b**6*c**2 + 20*a*b**8*c - b**10)) + 1/(2*c**3))*log(x + (-64*a**3*c**5*(-b*sq
rt(-(4*a*c - b**2)**5)*(30*a**2*c**2 - 10*a*b**2*c + b**4)/(2*c**3*(1024*a**5*c**5 - 1280*a**4*b**2*c**4 + 640
*a**3*b**4*c**3 - 160*a**2*b**6*c**2 + 20*a*b**8*c - b**10)) + 1/(2*c**3)) + 32*a**3*c**2 + 48*a**2*b**2*c**4*
(-b*sqrt(-(4*a*c - b**2)**5)*(30*a**2*c**2 - 10*a*b**2*c + b**4)/(2*c**3*(1024*a**5*c**5 - 1280*a**4*b**2*c**4
+ 640*a**3*b**4*c**3 - 160*a**2*b**6*c**2 + 20*a*b**8*c - b**10)) + 1/(2*c**3)) - 9*a**2*b**2*c - 12*a*b**4*c
**3*(-b*sqrt(-(4*a*c - b**2)**5)*(30*a**2*c**2 - 10*a*b**2*c + b**4)/(2*c**3*(1024*a**5*c**5 - 1280*a**4*b**2*
c**4 + 640*a**3*b**4*c**3 - 160*a**2*b**6*c**2 + 20*a*b**8*c - b**10)) + 1/(2*c**3)) + a*b**4 + b**6*c**2*(-b*
sqrt(-(4*a*c - b**2)**5)*(30*a**2*c**2 - 10*a*b**2*c + b**4)/(2*c**3*(1024*a**5*c**5 - 1280*a**4*b**2*c**4 + 6
40*a**3*b**4*c**3 - 160*a**2*b**6*c**2 + 20*a*b**8*c - b**10)) + 1/(2*c**3)))/(30*a**2*b*c**2 - 10*a*b**3*c +
b**5)) + (b*sqrt(-(4*a*c - b**2)**5)*(30*a**2*c**2 - 10*a*b**2*c + b**4)/(2*c**3*(1024*a**5*c**5 - 1280*a**4*b
**2*c**4 + 640*a**3*b**4*c**3 - 160*a**2*b**6*c**2 + 20*a*b**8*c - b**10)) + 1/(2*c**3))*log(x + (-64*a**3*c**
5*(b*sqrt(-(4*a*c - b**2)**5)*(30*a**2*c**2 - 10*a*b**2*c + b**4)/(2*c**3*(1024*a**5*c**5 - 1280*a**4*b**2*c**
4 + 640*a**3*b**4*c**3 - 160*a**2*b**6*c**2 + 20*a*b**8*c - b**10)) + 1/(2*c**3)) + 32*a**3*c**2 + 48*a**2*b**
2*c**4*(b*sqrt(-(4*a*c - b**2)**5)*(30*a**2*c**2 - 10*a*b**2*c + b**4)/(2*c**3*(1024*a**5*c**5 - 1280*a**4*b**
2*c**4 + 640*a**3*b**4*c**3 - 160*a**2*b**6*c**2 + 20*a*b**8*c - b**10)) + 1/(2*c**3)) - 9*a**2*b**2*c - 12*a*
b**4*c**3*(b*sqrt(-(4*a*c - b**2)**5)*(30*a**2*c**2 - 10*a*b**2*c + b**4)/(2*c**3*(1024*a**5*c**5 - 1280*a**4*
b**2*c**4 + 640*a**3*b**4*c**3 - 160*a**2*b**6*c**2 + 20*a*b**8*c - b**10)) + 1/(2*c**3)) + a*b**4 + b**6*c**2
*(b*sqrt(-(4*a*c - b**2)**5)*(30*a**2*c**2 - 10*a*b**2*c + b**4)/(2*c**3*(1024*a**5*c**5 - 1280*a**4*b**2*c**4
+ 640*a**3*b**4*c**3 - 160*a**2*b**6*c**2 + 20*a*b**8*c - b**10)) + 1/(2*c**3)))/(30*a**2*b*c**2 - 10*a*b**3*
c + b**5)) + (24*a**4*c**2 - 21*a**3*b**2*c + 3*a**2*b**4 + x**3*(50*a**2*b*c**3 - 30*a*b**3*c**2 + 4*b**5*c)
+ x**2*(32*a**3*c**3 + 11*a**2*b**2*c**2 - 19*a*b**4*c + 3*b**6) + x*(62*a**3*b*c**2 - 44*a**2*b**3*c + 6*a*b*
*5))/(32*a**4*c**5 - 16*a**3*b**2*c**4 + 2*a**2*b**4*c**3 + x**4*(32*a**2*c**7 - 16*a*b**2*c**6 + 2*b**4*c**5)
+ x**3*(64*a**2*b*c**6 - 32*a*b**3*c**5 + 4*b**5*c**4) + x**2*(64*a**3*c**6 - 12*a*b**4*c**4 + 2*b**6*c**3) +
x*(64*a**3*b*c**5 - 32*a**2*b**3*c**4 + 4*a*b**5*c**3))

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Giac [A]  time = 1.13125, size = 331, normalized size = 1.74 \begin{align*} -\frac{{\left (b^{5} - 10 \, a b^{3} c + 30 \, a^{2} b c^{2}\right )} \arctan \left (\frac{2 \, c x + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{{\left (b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}\right )} \sqrt{-b^{2} + 4 \, a c}} + \frac{\log \left (c x^{2} + b x + a\right )}{2 \, c^{3}} + \frac{3 \, a^{2} b^{4} - 21 \, a^{3} b^{2} c + 24 \, a^{4} c^{2} + 2 \,{\left (2 \, b^{5} c - 15 \, a b^{3} c^{2} + 25 \, a^{2} b c^{3}\right )} x^{3} +{\left (3 \, b^{6} - 19 \, a b^{4} c + 11 \, a^{2} b^{2} c^{2} + 32 \, a^{3} c^{3}\right )} x^{2} + 2 \,{\left (3 \, a b^{5} - 22 \, a^{2} b^{3} c + 31 \, a^{3} b c^{2}\right )} x}{2 \,{\left (c x^{2} + b x + a\right )}^{2}{\left (b^{2} - 4 \, a c\right )}^{2} c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c+a/x^2+b/x)^3/x,x, algorithm="giac")

[Out]

-(b^5 - 10*a*b^3*c + 30*a^2*b*c^2)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/((b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5
)*sqrt(-b^2 + 4*a*c)) + 1/2*log(c*x^2 + b*x + a)/c^3 + 1/2*(3*a^2*b^4 - 21*a^3*b^2*c + 24*a^4*c^2 + 2*(2*b^5*c
- 15*a*b^3*c^2 + 25*a^2*b*c^3)*x^3 + (3*b^6 - 19*a*b^4*c + 11*a^2*b^2*c^2 + 32*a^3*c^3)*x^2 + 2*(3*a*b^5 - 22
*a^2*b^3*c + 31*a^3*b*c^2)*x)/((c*x^2 + b*x + a)^2*(b^2 - 4*a*c)^2*c^3)